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Grade: 12th pass

                        

A body moving with uniform retardation covers 3km before it`s speed is reduced to half of its initial value it comes to rest in another distance of ?

4 years ago

Answers : (2)

Vikas TU
12286 Points
							
using 3rd law of eqn.
(u/2)^2 = u^2 – 2*a*3000
-3u^2/4 = – 6000a
u^2 = 8000a...................(1)
when it comes to rest,
0^2 = (u/2)^2 – 2as
2as = u^2/4
as = u^2/8
or
s = u^2/8a..............(2)
put eqn. (1) in eqn. (2)
and get s value.
4 years ago
Ankit Jaiswal
165 Points
							
let us suppose that its initial velocity = u
using equation of motion
v2=u2+2as
 
    (u/2)2 = u2 =2*a*3000 (value of final velocity is haif he initial taht is u/2 and s = 3 km =3000m)
\therefore (u2/4) – u=2*a*3000
\therefore (-3u2/4)=2*a*3000
\therefore – u2=(2*a*3000*4)/3
\therefore u= – 8000*a                      ….............................1
by appling the same equation when it comes to rest we get 
    02=(u/2)2 +2*a*s
\therefore – u2/4 =2*a*s
\therefore – u2/(4*2*a) = s
\therefore – u2/8a = s
after substituting values form 1 we get
\therefore – (-8000*a)/8a = s
\therefore s = 1000m = 1km
i think this might help 
 
plz approve if u found it useful 
thanks
 
 
 
4 years ago
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